求不定积分∫xarctanxdx

2024-12-04 09:32:13
推荐回答(2个)
回答1:

∫ x * arctanx dx

用分布积分法 令u=arctanx,v=x ,则∫ udv=uv-∫ vdu
=(1/2)∫ arctanxd(x²)

= ∫ arctanx d(x²/2)
= (x²/2)arctanx - (1/2)∫ x² d(arctanx)
= (x²/2)arctanx - (1/2)∫ x²/(x² + 1) dx
= (x²/2)arctanx - (1/2)∫ (x² + 1 - 1)/(x² + 1) dx
= (x²/2)arctanx - (1/2)∫ dx + (1/2)∫ dx/(x² + 1)
= (x²/2)arctanx - x/2 + (1/2)arctanx + C

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回答2:

满意请采纳,谢谢